Correct Statements Regarding Electrons in Atomic Models

To determine the correct statement(s), let us analyze each option individually:

1. Option (A): Heisenberg's Uncertainty Principle states that: $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$ To define a precise path or trajectory of a particle, both its exact position and exact velocity (or momentum) must be known simultaneously at any given instant. Since it is impossible to determine both of these quantities simultaneously with arbitrary precision for an electron, the concept of a definite path or trajectory is ruled out. Thus, statement (A) is correct.

2. Option (B): By convention, the energy of an electron at an infinite distance from the nucleus is taken as zero ($E_\infty = 0$), which corresponds to a free electron at rest. When an electron is bound in an atomic orbital (such as $2s$), it experiences electrostatic attraction towards the positively charged nucleus, resulting in a negative total energy ($E_{2s} < 0$). Since $E_{2s} < E_\infty$, the energy of the electron in the $2s$ orbital is lower than that of an electron infinitely far away. Thus, statement (B) is correct.

3. Option (C): According to Bohr's model, the energy of an electron in the $n^{\text{th}}$ orbit for a hydrogen-like species is given by: $E_n = -\frac{13.6\, Z^2}{n^2}\text{ eV}$ For $n = 1$: $E_1 = -13.6\, Z^2\text{ eV}$ This is the lowest (most negative) energy value possible, corresponding to the ground state, which represents the most stable orbit. Thus, statement (C) is correct.

4. Option (D): According to Bohr's model, the magnitude of the velocity of an electron in the $n^{\text{th}}$ orbit is given by: $v_n = \frac{2\pi k Z e^2}{n h} \propto \frac{1}{n}$ As $n$ increases, the magnitude of the orbital velocity decreases ($v_n \propto \frac{1}{n}$). Thus, statement (D) is incorrect.

Therefore, the correct statements are (A), (B), and (C).